Theorem sbalv 2206

Description: Quantify with new variable inside substitution. (Contributed by NM, 18-Aug-1993.)

Hypothesis

Ref	Expression
sbalv.1	⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)

Assertion

Ref	Expression
sbalv	⊢ ([𝑦 / 𝑥]∀𝑧𝜑 ↔ ∀𝑧𝜓)

Distinct variable groups:   𝑥,𝑧   𝑦,𝑧

Allowed substitution hints:   𝜑(𝑥, 𝑦, 𝑧)   𝜓(𝑥, 𝑦, 𝑧)

Proof of Theorem sbalv

Step	Hyp	Ref	Expression
1		sbal 2205	. 2 ⊢ ([𝑦 / 𝑥]∀𝑧𝜑 ↔ ∀𝑧[𝑦 / 𝑥]𝜑)
2		sbalv.1	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
3	2	albii 1851	. 2 ⊢ (∀𝑧[𝑦 / 𝑥]𝜑 ↔ ∀𝑧𝜓)
4	1, 3	bitri 278	1 ⊢ ([𝑦 / 𝑥]∀𝑧𝜑 ↔ ∀𝑧𝜓)

This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1567  [wsb 2098

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-11 2193

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-sb 2099

This theorem is used by:  sbex  2314  sbmo  2639  sbabel  2954  mo5f  33085